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Full text of "Treatise On Analysis Vol-Ii"

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take the values 0 and 00 simultaneously, it follows from ( that (b)
implies (c). Conversely, it is immediately seen that the function gf which is
equal to g~l on S and vanishes on X  S is //-measurable : for there exists a
partition of S (resp. X  S) into compact sets Ln (resp. Mn) and a //-negligible
set P (resp. Q) such that the restriction of g to each Ln is continuous, and it
follows that the restrictions of g' to Ln and Mn are continuous. Furthermore,
we have (fg)gf =fty$ , hence by the same argument it follows that (c) implies
(b). Hence it is enough to prove that (a) and (b) are equivalent.

Suppose first that/(ps is //-measurable. The set (JS is v-negligible, by virtue
of (1 3.14.1 .5) ; also the hypothesis implies the existence of a partition of S into
a sequence of compact sets LM and a /(-negligible set N, such that f\ Ln is
continuous for each n. Also N is v-negligible, by (; hence / is

Conversely, suppose that/is v-measurable. Then there exists a partition of
X consisting of a sequence of compact sets Hn and a v-negligible set N, such
that/I Hn is continuous for all n. Also there exists a partition of S consisting of
a //-negligible set L and a sequence of compact sets Kn such that g \ Kn is con-
tinuous (and >0) for all n. For all , we have inf g(x) = an > 0. Finally,


there exists also a partition of X  S into a /(-negligible set L' and a sequence
of compact sets K^ . It is clear that the restriction of fys to each of the
sets Hn n Km and K'm is continuous, and it is therefore enough to prove
that N n S is //-negligible. Now if //*(N n Krt) > 0, then we should have

^ an //*(N n Kn) > 0, which is absurd because


0 = V(N n S)



by virtue of (13.14.1). Hence each set N n Kn is //-negligible, therefore so
is N n S, and hence finally /<ps is //-measurable.

(13.14.3) Let g be a locally //-//?tegrable function which is ^0 on X, and let
v = g - //. Then a mapping f: X -> K is v-integrable if and only if (with the con-
ventions of (13.11))^ is \i-integr able \ and in that case we have

(                           \fdv= \fgdjA.

For/to be v-integrable it is necessary and sufficient that/4" and/"" should
be (13.7.4); and since we have(jfc)+ ~f+g, (fg)~ =/"#, with the conventions
of 13.11 about products, it is enough to prove (13.14.3) when/^ 0. The
first assertion is now a consequence of (13.14.1), (13.14.2), and (13.9.13); and
the relation ( is just ( w))~ |2,